Proof for mathd_algebra_513 - #200
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Proven lemmas: 1/1
The goal is to prove that if real numbers a, b ∈ ℝ satisfy the system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1.
This was decomposed into the two natural subgoals coming from the conjunction: prove a = 1 and prove b = 1. The proof uses the standard linear-system strategy: solve one variable from a + b = 2, substitute into 3a + 2b = 5, and then recover the other variable.
Progress is complete: 1 out of 1 theorem goals has been solved, and both component facts a = 1 and b = 1 are established. In Lean, this is handled very cleanly by splitting the conjunction and letting linarith solve each linear arithmetic subgoal directly from the two hypotheses.
Nothing remains to be proved for this statement. The main interesting feature is that the proof can be done either manually by substitution or automatically by linear arithmetic, and here the automatic linarith approach gives a short and robust proof.